Question: $f(x) = x+3$ $h(x) = 6x^{2}+x+2-2(f(x))$ $g(n) = -n^{2}-n+3(h(n))$ $ g(f(-7)) = {?} $
First, let's solve for the value of the inner function, $f(-7)$ . Then we'll know what to plug into the outer function. $f(-7) = -7+3$ $f(-7) = -4$ Now we know that $f(-7) = -4$ . Let's solve for $g(f(-7))$ , which is $g(-4)$ $g(-4) = -(-4)^{2}-(-4)+3(h(-4))$ To solve for the value of $g$ , we need to solve for the value of $h(-4)$ $h(-4) = 6(-4)^{2}-4+2-2(f(-4))$ To solve for the value of $h$ , we need to solve for the value of $f(-4)$ $f(-4) = -4+3$ $f(-4) = -1$ That means $h(-4) = 6(-4)^{2}-4+2+(-2)(-1)$ $h(-4) = 96$ That means $g(-4) = -(-4)^{2}-(-4)+(3)(96)$ $g(-4) = 276$